Who's done the math?

[bhsrobotics1671]

I was just curious..
Our team is in the middle of doing the math, trying to figure out the maximum force and acceleration that can be applied to the wheels before loosing traction.

Obviously it isn't going to be much, but we are trying to make a decision with our drive-train and gear box's.

So given the coefficients of friction..

From the Kit Of Parts Manual:

The tread material is Celcon M90, and has the following coefficients of friction on white, rippled fiberglass plastic sheet

Inline, static: 0.06
Inline, dynamic: 0.05
Transverse, static: 0.14
Transverse, dynamic: 0.10

So i guess what i am asking is has anyone figured out the maximum torque and acceleration that these wheels can handle on this surface?


Good Luck Teams!
THANKS!

[dmcguire3006]

Maximum force due to wheel friction f = umg (u = coef of friction)
Max force for 1 wheel = umg/4
Max torque T = Fr = umgr/4
For u = 0.05
mg = (50kg)(9.8m/s^2)= 500N
r = 0.05m
T = 1.25 N-m

Much less than the CIM can supply...

[bhsrobotics1671]

Thank You Very Much!!

I will share this with my team, hopefully it will help.


Good Luck!

[dbs12693x]

Originally Posted by dmcguire3006

Maximum force due to wheel friction f = umg (u = coef of friction)
Max force for 1 wheel = umg/4
Max torque T = Fr = umgr/4
For u = 0.05
mg = (50kg)(9.8m/s^2)= 500N
r = 0.05m
T = 1.25 N-m

Much less than the CIM can supply...

The coefficient used is improper. When wheels are rotating without sliding, use the coefficient of static friction to find the maximum friction between the wheel's outer edge and the point of contact. When the force pushing the wheel would make the friction required to roll higher than this theoretical maximum, the wheel slides.

It doesn't make much of a difference, though, in this example, and may have even been a typo. The coefficient used was .05, and the proper one per bhsrobotics1761 is .06.

The number is closer to .37, by my calculations, which is somehow much smaller and even less than dmcguire's number. I found the torque, by his numbers, to be .31 N-m.

In the end, your answer is "a very small number."

.

[bhsrobotics1671]

okay so in the end, the CIM is still powerful enough itself without a gearbox to accelerate, what is limiting you is the max velocity without the gearbox?

Thanks For Your Reply!

[NOV8R]

You guys need to check your assumptions. For example 50kg is only a 110 lb robot. I think a robot with battey and bumpers will be closer to 150 lbs or 68 kg. For a four wheel chassis with weight equally distributed each wheel would exert 2.25 lb (150x0.06/4) before breaking free. That equates to 0.56 ft-lbs or 108 oz-in of torque.

[bhsrobotics1671]

Very good point. We have been using 145lbs in most of our math.

[chris31]

Originally Posted by NOV8R

You guys need to check your assumptions. For example 50kg is only a 110 lb robot. I think a robot with battey and bumpers will be closer to 150 lbs or 68 kg. For a four wheel chassis with weight equally distributed each wheel would exert 2.25 lb (150x0.06/4) before breaking free. That equates to 0.56 ft-lbs or 108 oz-in of torque.

You should also calculate in the trailer you will be dragging behind you.

[dmcguire3006]

Sorry, arithmetic is not my forte... The equations are correct. I agree with the updated numbers - in any case much smaller than a CIM's max torque.

I did use the kinetic rather than tha static coefficient of friction. One of the secrets in this game will be to NOT to accelerate faster than allowed by the coefficient of friction, in this case 0.06g or .6m/s^2.

Thanks!

[bhsrobotics1671]

Originally Posted by chris31

You should also calculate in the trailer you will be dragging behind you.

Is this just mainly affecting the change in CG? or would there be another variable it would have an affect on?

[DavidGitz]

If you add more wheels to the robot that are powered, you would get more force right? So then here's a couple questions.

1) Why not use every motor in the kit powering as many wheels? Obviously not very feasible and a bunch of other reasons why not to do this, but what would the maximum force be that you could generate?

2) Or, just have your drive motors connected to more than 1 wheel, like a 4 wheel-drive system with 2 wheels (or more). What would the benefits be?

Just curious, I understand about the higher complexity/weight and all that stuff.

[NOV8R]

You're only going to get about 9 lbs of tractive force regardless of how many wheels you use. 150 lbs x 0.06 = 9 lbs. More wheels just divide it up.

[bhsrobotics1671]

Not directly related to the Friction stats now.. but more towards the battery power.

Some team members have brought up battery drainage issues if we were to go straight CIM to wheel.

Others have argued that the low resistance level will require less of an elec. current for the motors.

Our other concern points to the new electrical system. Does anybody know if this system will require more or less power than the old system? Our main concern is the wireless router...


Has anyone else brought up battery concerns? With our experience with the batteries, they seem to hold up pretty well.

[AndrewN]

Robot = 150lb (120lb + battery + bumpers)
Max force available to move robot: 0.06 * 150lb = 9lb
Trailer ~ 40lb (from eye witness reports)
Drag on robot from trailer: 0.06 * 40lb = 2.4lbs

Total: 9 - 2.4 = 6.6lbs force

Total mass to move forward: 190lbs

I'll leave the acceleration calculations for you ...
Now tell me the top speed from rest when you slam into the wall after 50ft?
What's the maximum speed before you need to start braking so you don't crash?
How long did the traversal of the whole field take?

While braking, did you account for the center of mass moving forward over the front wheels? If so how fast could you brake? What happens if the trailer jack-knifes into the robot while braking?

So many questions, so little time!

[bhsrobotics1671]

tell me about it. thanks for your help. i will be bringing this up with my team tomorrow.

this will definitely help with our calculations, i hope we will be able to put it all together and make a decision.

thanks again!